ba 452 lesson c.2 waiting line simulation 1 readingsreadings chapter 12 simulation
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BA 452 Lesson C.2 Waiting Line Simulation 1 1
Readings
Readings
Chapter 12Simulation
BA 452 Lesson C.2 Waiting Line Simulation 2 2
Overview
Overview
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Overview
Bidding Simulations estimate the probability that various bids will win, then performs Risk Analysis. The simplest analysis determines the bid that maximizes expected gains.
Product Failure Simulations estimate the probability that a product fails after various amounts of use, then performs risk analysis, trading off consumer confidence of warranties with expected costs.
Waiting Line Simulations model queuing when there are no analytical formula, such as non-Markov arrival distributions or service-time distributions.
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Tool Summary Use the Excel 2003 workbook general-purpose template http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/Template.xls Simulate Bidding with rival buyers or sellers. If buyers, the
high bid wins; if sellers, the low bid. In either case, the objective is uncertain, so maximize expected value.
Use Summary Statistics to simulate the probability of product failure.
Simulate Single Channel Waiting Lines with the specialized Excel 2003 workbook template
http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/SingleWait.xls
Overview
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Bidding
Bidding
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Bidding
Overview
Bidding Simulations estimate the probability that various bids will win, then perform Risk Analysis. The simplest analysis determines the bid that maximizes expected gains.
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Bidding
Strassel Investors buys real estate, develops it, and resells it for a profit. A new property is available, and Bud Strassel believes [is certain] it can be sold for $160,000.
The current property owner asked for bids and stated that the property will be sold for the highest bid in excess of $100,000.
Two competitors will be submitting bids for the property. Strassel assumes, for planning purposes, that the bid by each competitor is [independent and] uniformly distributed between $100,000 and $150,000.
Simulate the probability that Strassel can win with a bid of $130,000. Use a 500 trial simulation.
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The bid by each competitor is [independent and] uniformly distributed between $100,000 and $150,000.
Strassel must beat the maximum of the two bids.
Bidding
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Count the number of times out of 500 that the $130,000 bid is better than the maximum bid by competitors.
Then, compute the frequency (probability) of outbidding competitors.
Bidding
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Compare that probability of 0.3400 for a $130,000 bid with probabilities for $140,000 and $150,000 bids.
It is better to compute both probabilities on a single simulation, rather than running two separate simulations (as on the right).
Bidding
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Compute expected profit for each bid. Profit ($130,000 bid) = 0.3400 x ($160,000-$130,000) =
$10,200. Profit ($140,000 bid) = 0.6260 x ($160,000-$140,000) =
$12,520. Profit ($150,000 bid) = 1.0000 x ($160,000-$150,000) =
$10,000. So, bid $140,000, and loose some of the time. It is more accurate to compute all of those probabilities
from a single simulation, rather than from 3 separate simulations.
Bidding
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Product Failure
Product Failure
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Overview
Product Failure Simulations estimate the probability that a product fails after various amounts of use, then performs Risk Analysis, trading off the gains from consumer confidence of offering warranties with the expected cost.
Product Failure
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Goodyear Tire Company has produced a new tire with an estimated mean lifetime mileage of 36,500 miles. Management also believes that the standard deviation is 5000 miles, and mileage is normally distributed. [Of course, negative miles make no sense, but that is unlikely since it negative values of more than 7 standard deviations from the mean.]
Simulate the probability that a tire lasts more than 40,000 miles; more than 32,000 miles; more than 30,000 miles; and more than 28,000 miles.
What tire mileage guarantee do you recommend if management wants no more than 10% of tires to fail the guaranteed amount?
Product Failure
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… a new tire with an estimated mean lifetime mileage of 36,500 miles … standard deviation is 5000 miles, and mileage is normally distributed.
Product Failure
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Simulate the probability that a tire lasts more than 40,000 miles; more than 32,000 miles; more than 30,000 miles; and more than 28,000 miles.
What tire mileage guarantee do you recommend if management wants no more than 10% of tires to fail the guaranteed amount? 30,000 miles has 9% failure.
Product Failure
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Waiting Line
Waiting Line
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Overview
Waiting Line Simulations model queuing when there are no analytical formula, such as non-Markov arrival distributions (non-Poisson) or non-Markov service-time distributions (non-exponential).
Waiting Line
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For the next example, download Excel 2003 workbook http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/SingleWait.xls
Waiting Line
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Skandinaviska Enskilda Banken AB (SEB) will open several new branch banks during the coming year. Each new branch is designed to have one automated teller machine (ATM). A concern is that during busy periods several customers may have to wait to use the ATM. This concern prompted the bank to study the ATM waiting line system. The bank established the guideline that the average customer waiting time should be 1 minute or less.
Waiting Line
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One probabilistic input to the ATM simulation model is the arrival times of customers who use the ATM. In waiting line simulations, arrival times are determined by randomly generating the time between two successive arrivals. For the SEB ATM system, the interarrival times are continuously uniformly distributed between 0 and 5 minutes.
Waiting Line
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The other probabilistic input is the service time, which is the time a customer spends using the ATM machine. Past data indicates a normal probability distribution with a mean of 2 minutes and a standard deviation of 0.5 minutes. (Although negative service times will be generated by the normal distribution, they are more than 4 standard deviations from the mean, and so have an insignificant effect on the simulation.)
Waiting Line
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All formulae for Customers 3 through 1000 are copied from the second customer.
Results for Customers 6 through 995 are hidden below.
Waiting Line
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Summary Statistics are computed for the steady state of the system.
Results are only used for Customers 101 through 1000 because Customers 1 through 100 face unusually short lines for the ATM.
The steady state is relevant for an ATM open 24 hours per day. (Of course, using the same interarrival distribution for 4am as for 4pm is then inaccurate.)
Waiting Line
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Statistics for Customers 100, … use:
Waiting Time (E116, …, E1015)
Service Time (F116, …, F1015)
Completion Time (G116, …, G1015)
Number Waiting counts the number of customers with positive Waiting Times.
Utilization of ATM sums Service Times and divides that sum by the time between when Customer 100 could have been served (G115) to when Customer 1000 has completed service (G1015).
Waiting Line
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The central limit theorem predicts the Average Waiting Time from many repeated simulations is most likely close to the expected value (mean) of waiting times for the true probability distribution of waiting times.
The expected value of waiting times is greater than 1 minute. Therefore: 1 ATM is insufficient for the bank to meet satisfy the
guideline that the average customer waiting time should be 1 minute or less.
Waiting Line
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Review Questions
Review Questions You should try to answer some of the following
questions before the next class. You will not turn in your answers, but students may
request to discuss their answers to begin the next class. Your upcoming Final Exam will contain some similar
questions, so you should eventually consider every review question before taking your exams.
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Review 1: Waiting Lines
Review 1: Waiting Lines
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For the next example, download Excel 2003 workbook http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/SingleWaitInNOut.xls
Review 1: Waiting Lines
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A fast-food franchise has received several complaints about long waits for service at its Newbury Park location.
The franchise is especially concerned about waits of over 5 minutes.
Hence, the franchise collects data on interarrival times of customers, and service times.
Show how to use that data to estimate the probability of a customer waiting over 5 minutes for service.
Review 1: Waiting Lines
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Assume the interarrival times between customers at its Newbury Park location are estimated to be continuously uniformly distributed between 0 and 7minutes.
Assume customer service times from workers of average abilities follow a finite distribution: With probability .20, the customer is chatty and pays in exact
change, and so takes 4 minutes to serve. With probability .30, the customer is chatty and pays with a
credit card, and so takes 3 minutes to serve. With probability .15, the customer is quiet and pays in exact
change, and so takes 2.5 minutes to serve. With probability .35, the customer is quiet and pays with a credit
card, and so takes 2 minutes to serve.
Review 1: Waiting Lines
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The Excel 2003 workbook http://faculty.pepperdine.edu/jburke2/ba452/PowerP2/SingleWaitInNOut.xls
constructs and explains the single-channel waiting line simulation.
Most estimates for the probability of a customer waiting over 5 minutes for service are between 10% and 18%.
Save that workbook to use as a template for other single-channel waiting line simulation that use other distributions for interarrival times or service times.
Review 1: Waiting Lines
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BA 452 Quantitative Analysis
End of Lesson C.2